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Let us denote the residue class of $latex p$ by $latex \mathbb{Z}_p$. Show that, there exist non-zero $latex b,c$ in $latex \mathbb{Z}_p$ such that $latex y\equiv bx\;\text{mod}\; p$ and $latex z\equiv cy\;\text{mod}\; p$.

From hint 2, show that $latex (1+b+bc)^2\equiv ab^2cx\;\text{mod}\;p$. This means that $latex x\equiv a^{-1}c^{-1}(1+b^{-1}+c)^2\;\text{mod}\;p$ (you need to convince yourself that the inverses exist).  Now it becomes a matter of simply choosing $latex b,c$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]

Note that, $latex 1+b^{-1}+c$ cannot be zero. Given any $latex b\neq p-1$, there exists exactly one non-zero $latex c$ such that $latex 1+b^{-1}+c$ is 0 modulo $latex p$. Hence, in this case there are $latex (p-2)^2$ choices. For $latex b=p-1$, this special $latex c$ is actually 0. Hence in this case there are $latex p-1$ choices. Thus, the total number of choices is $latex (p-2)^2+(p-1)=p^2-4p+4+(p-1)=p^2-3p+3$. Adding to this the $latex 3p-1$ cases considered in hint 1, we get $latex p^2+1$ as the answer.

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